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Let $\{X_n\}_{n = 1}$ be a sequence of random variables.

Let $Y_n = \sum_{i=1}^n X_i$

Then, I want to show that $\sigma(Y_1,Y_2,\dots,Y_n) = \sigma(X_1,X_2,\dots,X_n)$ for every $n$.

It is clear to me that $\sigma(X_1,X_2,\dots,X_n) \subset \sigma(Y_1,Y_2,\dots,Y_n)$. But, how do I show the inclusion in the other direction?

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    $\begingroup$ $X_n = Y_n - Y_{n-1}$. $\endgroup$ – Stephen Montgomery-Smith Aug 22 '15 at 23:14
  • $\begingroup$ Hint: If $(Y_1, \dots, Y_n) = f(X_1,\dots,X_n) = (f_1(X_1, \dots, X_n),\dots,f_n(X_1,\dots,X_n))$, then $\sigma(Y_1, \dots, Y_n) \subset \sigma(X_1,\dots,X_n)$. For one direction, $f_i = \sum_{k=1}^i X_k$. What about $f^{-1}$? $\endgroup$ – snar Aug 22 '15 at 23:16
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This says

  • The occurrence or non-occurrence of every event whose occurrence or non-occurrence is determined by the values of $X_1,\ldots,X_n$ is determined by the values of $Y_1,\ldots,Y_n$; and
  • The occurrence or non-occurrence of every event whose occurrence or non-occurrence is determined by the values of $Y_1,\ldots,Y_n$ is determined by the values of $X_1,\ldots,X_n$.

Most of the work of proving this is accomplished by showing that (putting it informally) if you know the values of $X_1,\ldots,X_n$ you can find $Y_1,\ldots,Y_n$ and if you know the values of $Y_1,\ldots,Y_n$ you can find $X_1,\ldots,X_n$.

Somewhat more precisely: for every $n$-tuple of values of $X_1,\ldots,X_n$ there is only one corresponding $n$-tuple of values of $Y_1,\ldots,Y_n$, and vice-versa.

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